1. The problem statement, all variables and given/known data A non uniform rod of length L=1m and linear mass density: λ(x)=λo(1-(x/2L)) with λo=1Kg/m. It is supported at it's midpoint and initially tips over so that it's left end lies on the ground. Part a: Find the position of the centre of mass along the rod when it is in it's horizontal position. Part b: One begins to fix small uniform metal balls of radius r=1cm and mass m=50g to the lighter end of the unbalanced rod. Starting from the end and moving inwards with each ball touching the next. How many balls does it take to tip the rod over? That is the bones of the question. Here is a link to the actual question more clearly outlined with diagrams. http://i.imgur.com/Hnr3z.png 2. Relevant equations M=∫λ(x)dx X(centre of mass position)= 1/m∫xλ(x)dx 3. The attempt at a solution I have already solved part a. The mass of the rod is 0.75Kg and the position of the centre of mass (Xcm) is 4/9M (0.44444m). I have tried for hours today and yesterday to solve part b but not had any luck. My original idea was to make the centre of mass L/2 (0.5m) and then solve for the mass. Then I got the difference in the masses and tried to find how many metal balls would equal that mass. This didn't feel right as the size and position of the balls of the rod are not been taken into account. I don't know how I am meant to approach this question at all. Is it a case of balancing masses, of balancing the forces on either side or perhaps splitting the rod into segments that make it easier to deal with? Any help/suggestions would be greatly appreciated.